Spontaneous processes such as neutron decay require that the final state is lower in energy than the initial state. In (stable) nuclei, this is not the case, because the energy you gain from the neutron decay is lower than the energy it costs you to have an additional proton in the core.
For neutron decay in the nuclei to be energetically favorable, the energy gained by the decay must be larger than the energy cost of adding that proton. This generally happens in neutron-rich isotopes:
An example is the $\beta^-$-decay of Cesium:
$$\phantom{Cs}^{137}_{55} \mathrm{Cs} \rightarrow \vphantom{Ba}^{137}_{56}\mathrm{Ba} + e^- + \bar{\nu}_e$$
For a first impression of the energies involved, you can consult the semi-empirical Bethe-Weizsäcker formula which lets you plug in the number of protons and neutrons and tells you the binding energy of the nucleus. By comparing the energies of two nuclei related via the $\beta^-$-decay you can tell whether or not this process should be possible.
Protons and neutrons are referred to collectively as nucleons. Nucleons interact via the strong nuclear force, and this interaction can't be expressed by any simple equation. The reason is that nucleons are not fundamental particles. They're actually clusters of quarks.
Short range
The low-energy structure of nuclei is amazingly insensitive to the details of the nucleon-nucleon interaction that you pick as an approximation to the actual underlying quark-quark interaction. This is both good and bad. It's good because you don't need to understand very much about the nasty details in order to find out the properties of nuclei, e.g., why nuclei have the sizes and shapes they do. It's bad because it means that you can never infer very much about the interaction simply by observing properties of nuclei. As an example of how insensitive nuclear structure is to the details of the strong nuclear force, clusters of sodium atoms have magic numbers that match up with the first few magic numbers for nuclei; this is because these magic numbers only depend on the short-range nature of the interaction.
Other effects that can be understood based on the short range of the interaction are:
Nuclei act as though they have surface tension (so they resist being deformed).
Nuclei are most stable if they have even numbers of neutrons and even numbers of protons (because then the neutrons and protons can pair off in time-reversed orbits that maximize their spatial overlap).
Nucleons in an open shell tend to couple so as to form the minimum total angular momentum (the opposite of Hund's rules for electrons).
A residual interaction
The short-range nature of the nuclear interaction is very surprising, because the quark-quark force is believed to be roughly independent of distance. What's happening here is that nucleons are color-neutral, just as a hydrogen atom is charge-neutral. Just as hydrogen atoms "shouldn't" interact, nucleons "shouldn't" interact either. The forces between nucleons very nearly cancel out, and likewise the electrical forces between two neutral hydrogen atoms very nearly cancel out. The nonvanishing interaction comes from effects like the polarization of one particle by the other. For this reason, the nucleon-nucleon interaction is referred to as a residual interaction.
Other than its coupling constant and its range, what other features of the nuclear interaction are important for understanding low-energy nuclear structure?
Spin-orbit
There is a spin-orbit interaction, which is much stronger than, and in the opposite direction compared to, the one expected from special relativity alone.
Symmetry between neutrons and protons
The nuclear interaction remains unchanged when we transform neutrons to protons and protons to neutrons. For this reason light nuclei exhibit nearly identical properties if you swap their N and Z. Heavy nuclei don't have this symmetry, which is broken by the electrical interaction.
No qualitative features inferrable from sizes of nuclei
We do not get any clearcut, qualitative information about the interaction based on the observed sizes of nuclei. An extremely broad class of interactions between point particles results in n-body systems that have bound states and finite density. The finite density (i.e., the lack of a total collapse to a point) is essentially a generic result of the Heisenberg uncertainty principle. Only for certain special types of potentials that blow up to $-\infty$ at short ranges can one circumvent this (Lieb 1976).
A variety of models
Because the nucleon-nucleon interaction is a residual interaction, and nucleons are really composite rather than pointlike, the whole notion of a nucleon-nucleon interaction is an approximation, and one can model it in a variety of ways while still producing agreement with the data. In particular, some models have a hard, repulsive core, while others do not,(Chamel 2010, Stone 2006) and both types can reproduce the observed sizes of nuclei. This disproves the common misconception that such a hard core is needed in order to explain the sizes of nuclei.
References
Chamel and Pearson, 2010, "The Skyrme-Hartree-Fock-Bogoliubov method: its application to finite nuclei and neutron-star crusts," http://arxiv.org/abs/1001.5377
Lieb, Rev Mod Phys 48 (1976) 553, available at http://www.pas.rochester.edu/~rajeev/phy246/lieb.pdf
Stone and Reinhard, 2006, "The Skyrme Interaction in finite nuclei and nuclear matter," http://arxiv.org/abs/nucl-th/0607002
Best Answer
In a nucleus whose N/Z ratio is too large, the Pauli exclusion principle forces many of the neutrons to be in states with high energies. This makes the system less stable. For a fixed N, adding protons also makes such highly neutron-rich systems more stable, because the interaction between the protons and the neutrons is attractive, and the protons can go into low-energy states.
There is no Coulomb barrier for neutrons, so if a neutron has a high enough energy to escape, it just does -- no tunneling is required. Even if the system is bound, the system undergoes beta decay toward the line of stability.
There are nevertheless some systems with very high N/Z that are stable against neutron emission. E.g., 8He is bound and has a half-life of 119 ms.
The two pure-neutron systems that theorists predict might have the best chances of holding together are N=2 (the dineutron) and N=4. Experimental searches for dineutrons over a period of decades have failed to find any, so we're pretty sure they're unbound. These people claimed to have detected the N=4 system, with a lifetime of at least ~100 ns, which means it would have to be bound, although not stable with respect to beta decay. Whether they're right is a whole different question. I wouldn't bet a six-pack on it.